A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Equivalently, if a particle travels in a closed loop, the net work done (the sum of the force acting along the path multiplied by the distance travelled) by a conservative force is zero.

Does the work done by the force remain 0 even if it varies at all points on the loop? From the definition given in wikipedia it seems as if it is defined for work done by a constant conservative force. Is the work done over a loop 0 for variable forces as well? Can it be somehow proven for a variable force?


Does the work done by the force remain 0 even if it varies at all points on the loop ?

Yes. For example, the gravitational force.

Note, that in general fields are not conservative. So if you write an arbitrary force, the work will not be zero.

  • $\begingroup$ No I mean varying with point and for gravitational force we can take approximation that it is constant. I know the work is 0 even for forces varying extremely from point to point over a loop too, but can we just apply it everywhere ? How ? $\endgroup$ – Rijul Gupta Nov 23 '13 at 18:31
  • $\begingroup$ physics.stackexchange.com/questions/87621/… please take a look and tell me what I am doing wrong here then ? $\endgroup$ – Rijul Gupta Nov 23 '13 at 18:48
  • $\begingroup$ The field is electrostatic, I think it should be conservative only $\endgroup$ – Rijul Gupta Nov 23 '13 at 19:11
  • $\begingroup$ So if I set up an electric field which has a dependency on y as I mentioned then in that case it is possible that the field may not be conservative even though it is electrostatic in nature ? $\endgroup$ – Rijul Gupta Nov 23 '13 at 19:32
  • $\begingroup$ But just in case hypothetically, I did what will be its consequence ? $\endgroup$ – Rijul Gupta Nov 23 '13 at 19:40

The work done by a force along a path $\gamma$ is defined as $$ W = \int_\gamma \vec{F}(\vec{r})\cdot d\vec{r} = \int_a^b \vec{F}(\gamma(t))\cdot\dot{\gamma}(t)\,dt$$ where the last equality is the actual definition of the line-integral of a vectorfield along a curve. Note that the force is explicitly depending on position. As you state correctly, a force is called conservative iff the work done is a function of start- and end-point of the path only: $$ F\, cons. \Leftrightarrow W = V[\gamma(a)]-V[\gamma(b)] \Leftrightarrow F=-\nabla V \Rightarrow \nabla\times F=0$$

The last implication is an equivalence if the domain on which $F$ is defined, is star-shaped. It is a special case of the Poincaré Lemma.

Maxwells equations tell us, that in the absence of time-variing magnetic fields, the electric field is curl-free and hence posesses a potential $$\nabla\times E=0 \Rightarrow E=-\nabla V$$ Hence every electrostatic field is conservative, which can easily be checked by calculating the curl!

Sidenote (Please correct me if i that is not on spot):
You may argue, that the electrostatic field of a point-charge $E=-q\,\hat{r}/r^2$ is singular at the origin and the curl-criterion therfore not strictly applicable. However this is kind of an artifact from modeling the chargedistribution as a Dirac Delta $\rho=q\,\delta(\vec{r})$ infinitly narrow peaked in space. One may instead think of 'smearing out' the singular distribution a bit in space, so everything becomes nice and smooth, making Poincarés Lemma applicable.


protected by Qmechanic Dec 28 '16 at 6:05

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.